• Chi-squared test

segmentation |> count(Segment)

# A tibble: 4 × 2
  Segment        n
  <chr>      <int>
1 Moving up     70
2 Suburb mix   100
3 Travelers     80
4 Urban hip     50

segmentation |> 
  count(subscribe, ownHome) |> 
  pivot_wider(id_cols = subscribe,
              names_from = ownHome,
              values_from = n)

# A tibble: 2 × 3
  subscribe ownNo ownYes
  <chr>     <int>  <int>
1 subNo       137    123
2 subYes       22     18

• Chi-squared test for given probabilities

\(H_0: p_1 = \frac{1}{4} \land p_2 = \frac{1}{4} \land p_3 = \frac{1}{4} \land p_4 = \frac{1}{4}\)

\(H_1: p_1 \neq \frac{1}{4} \lor p_2 \neq \frac{1}{4} \lor p_3 = \frac{1}{4} \lor p_4 \neq \frac{1}{4}\)

\(\begin{align} \chi^2 & = \sum_{i=1}^n \frac{(Observed_i - Expected_i)^2}{Expected_i} \\ & = \frac{(70 - 300\frac{1}{4})^2}{300\frac{1}{4}} + \frac{(100 - 300\frac{1}{4})^2}{300\frac{1}{4}} + \frac{(80 - 300\frac{1}{4})^2}{300\frac{1}{4}} + \frac{(50 - 300\frac{1}{4})^2}{300\frac{1}{4}} \end{align}\)

• Chi-squared test for given probabilities

\(H_0: p_1 = \frac{1}{4} \land p_2 = \frac{1}{4} \land p_3 = \frac{1}{4} \land p_4 = \frac{1}{4}\)

\(H_1: p_1 \neq \frac{1}{4} \lor p_2 \neq \frac{1}{4} \lor p_3 = \frac{1}{4} \lor p_4 \neq \frac{1}{4}\)

\(\begin{align} \chi^2 & = \sum_{i=1}^n \frac{(Observed_i - Expected_i)^2}{Expected_i} \\ & = \frac{(70 - 300\frac{1}{4})^2}{300\frac{1}{4}} + \frac{(100 - 300\frac{1}{4})^2}{300\frac{1}{4}} + \frac{(80 - 300\frac{1}{4})^2}{300\frac{1}{4}} + \frac{(50 - 300\frac{1}{4})^2}{300\frac{1}{4}} \end{align}\)

• Pearson’s Chi-squared test

\(H_0: p_{11} = \frac{260}{300}\frac{159}{300} \land p_{12} = \frac{260}{300}\frac{141}{300} \land p_{21} = \frac{40}{300}\frac{159}{300} \land p_{22} = \frac{40}{300}\frac{141}{300}\)

\(H_1: p_{11} \neq \frac{260}{300}\frac{159}{300} \lor p_{12} \neq \frac{260}{300}\frac{141}{300} \lor p_{21} \neq \frac{40}{300}\frac{159}{300} \lor p_{22} \neq \frac{40}{300}\frac{141}{300}\)

\(\begin{align} \chi^2 & = \sum_{i=1}^n \frac{(Observed_i - Expected_i)^2}{Expected_i} \\ & = \frac{(137 - 300\frac{260}{300}\frac{159}{300})^2}{300\frac{260}{300}\frac{159}{300}} + \frac{(123 - 300\frac{260}{300}\frac{141}{300})^2}{300\frac{260}{300}\frac{141}{300}} \\ & \: \: \: + \frac{(22 - 300\frac{40}{300}\frac{159}{300})^2}{300\frac{40}{300}\frac{159}{300}} + \frac{(18 - 300\frac{40}{300}\frac{141}{300})^2}{300\frac{40}{300}\frac{141}{300}} \end{align}\)

• Pearson’s Chi-squared test

\(H_0: p_{11} = \frac{260}{300}\frac{159}{300} \land p_{12} = \frac{260}{300}\frac{141}{300} \land p_{21} = \frac{40}{300}\frac{159}{300} \land p_{22} = \frac{40}{300}\frac{141}{300}\)

\(H_1: p_{11} \neq \frac{260}{300}\frac{159}{300} \lor p_{12} \neq \frac{260}{300}\frac{141}{300} \lor p_{21} \neq \frac{40}{300}\frac{159}{300} \lor p_{22} \neq \frac{40}{300}\frac{141}{300}\)

\(\begin{align} \chi^2 & = \sum_{i=1}^n \frac{(Observed_i - Expected_i)^2}{Expected_i} \\ & = \frac{(137 - 300\frac{260}{300}\frac{159}{300})^2}{300\frac{260}{300}\frac{159}{300}} + \frac{(123 - 300\frac{260}{300}\frac{141}{300})^2}{300\frac{260}{300}\frac{141}{300}} \\ & \: \: \: + \frac{(22 - 300\frac{40}{300}\frac{159}{300})^2}{300\frac{40}{300}\frac{159}{300}} + \frac{(18 - 300\frac{40}{300}\frac{141}{300})^2}{300\frac{40}{300}\frac{141}{300}} \end{align}\)

• 2 sample t-test: independent samples

segmentation |> ggplot() + 
  geom_histogram(aes(x = income), color='black') + 
  facet_wrap(facets = vars(ownHome))

• 2 sample t-test: independent samples

\(H_0: \mu_{ownNo} - \mu_{ownYes}= 0\) \(H_1: \mu_{ownNo} - \mu_{ownYes} \neq 0\)

\(t = \frac{\overline{ownNo} - \overline{ownYes}}{\sqrt{\frac{s_{ownNo}^2}{n_{ownNo}} - \frac{s_{ownYes}^2}{n_{ownYes}}}} = \frac{47391.01 - 54934.68}{\sqrt{ \frac{358692875}{159} - \frac{430890091}{141}}} \approx -3.273094\)

• R base way

t_test <- t.test(income ~ ownHome, data = segmentation,
                 alternative='two.sided', mu = 0,
                 conf.level = 0.95)
t_test

    Welch Two Sample t-test

data:  income by ownHome
t = -3.2731, df = 285.25, p-value = 0.001195
alternative hypothesis: true difference in means between group ownNo and group ownYes is not equal to 0
95 percent confidence interval:
 -12080.155  -3007.193
sample estimates:
 mean in group ownNo mean in group ownYes 
            47391.01             54934.68 

• 2 sample t-test: independent samples

  • Confidence interval:

\[c_L < \mu_{ownNo} - \mu_{ownYes} < c_U\]

• \(\mu_{ownNo} - \mu_{ownYes}\) is not a random variable so we need to use a random variable

\[P \Biggr( t_L < \frac{\overline{x}_{ownNo} - \overline{x}_{ownYes} - (\mu_{ownNo} - \mu_{ownYes})}{\sqrt{\frac{s^2_{ownNo}}{n_{ownNo} } +\frac{s^2_{ownYes}}{n_{ownYes}}}} < t_U \Biggr) = 0.95\]

• \(\overline{x}_{ownNo} - \overline{x}_{ownYes}\) is a random variable

• 2 sample t-test: independent samples

  • Confidence interval:

    • \(\frac{\overline{x}_{ownNo} - \overline{x}_{ownYes} - (\mu_{ownNo} - \mu_{ownYes})}{\sqrt{\frac{s^2_{ownNo}}{n_{ownNo} } +\frac{s^2_{ownYes}}{n_{ownYes}}}}\) is also a random variable with student’s t-distribution and \(\nu \approx \frac{(\frac{s_{ownNo}^2}{n_{ownNo}} + \frac{s_2^2}{n_{ownYes}})^2}{\frac{(\frac{s_{ownNo}^2}{n_{ownNo}})^2}{n_{ownNo}-1} + \frac{(\frac{s_2^2}{n_{ownYes}})^2}{n_{ownYes}-1}} = 285.2521\) degrees of freedom

    • Also we need to specify \(t_L\) and \(t_U\)

t_L <- qt(p = 0.025, df = 285.25, lower.tail = TRUE)
t_L

[1] -1.968315

t_U <- qt(p = 0.975, df = 285.25, lower.tail = TRUE)
t_U

[1] 1.968315

• 2 sample t-test: independent samples

  • Confidence interval:

\[P(-7543.674 - 1.968315\times2304.753 \\ < \mu_{ownNo} - \mu_{ownYes} < -7543.674 - 1.968315\times2304.753) = 0.95\] \[P(-12080.16 < \mu_{ownNo} - \mu_{ownYes} < -3007.193) = 0.95\]

• In the long run 95% of confidence intervals constructed in this manner will contain the true parameter

• 2 sample t-test: independent samples

\(H_0: \mu_{ownNo} - \mu_{ownYes}= 0\)

\(H_1: \mu_{ownNo} - \mu_{ownYes} \neq 0\)

\(t = \frac{\overline{ownNo} - \overline{ownYes}}{\sqrt{\frac{s_{ownNo}^2}{n_{ownNo}} - \frac{s_{ownYes}^2}{n_{ownYes}}}} = \frac{47391.01 - 54934.68}{\sqrt{ \frac{358692875}{159} - \frac{430890091}{141}}} \approx -3.273094\)

• tidymodels way

segmentation |> 
  t_test(formula = income ~ ownHome,
         alternative = "two-sided",
         order = c("ownNo", "ownYes"),
         mu = 0,
         conf_level = 0.95)

# A tibble: 1 × 7
  statistic  t_df p_value alternative estimate lower_ci upper_ci
      <dbl> <dbl>   <dbl> <chr>          <dbl>    <dbl>    <dbl>
1     -3.27  285. 0.00119 two.sided     -7544.  -12080.   -3007.

• Testing Multiple Group Means: Analysis of Variance (ANOVA)

segmentation |> 
  group_by(Segment) |> 
  summarise(mean = mean(income),
            variance = var(income),
            n = n())

# A tibble: 4 × 4
  Segment      mean   variance     n
  <chr>       <dbl>      <dbl> <int>
1 Moving up  53091.  92862689.    70
2 Suburb mix 55034. 142761527.   100
3 Travelers  62214. 564173979.    80
4 Urban hip  21682.  23885953.    50

• Testing Multiple Group Means: Analysis of Variance (ANOVA)

segmentation |> 
  distinct(Segment) |> 
  arrange(Segment) |> 
  rowid_to_column(var = 'i')

# A tibble: 4 × 2
      i Segment   
  <int> <chr>     
1     1 Moving up 
2     2 Suburb mix
3     3 Travelers 
4     4 Urban hip 

segmentation |> 
  distinct(ownHome) |> 
  rowid_to_column(var = 'j')

# A tibble: 2 × 2
      j ownHome
  <int> <chr>  
1     1 ownNo  
2     2 ownYes 

• Testing Multiple Group Means: Analysis of Variance (ANOVA)

mu_ij <- segmentation |> 
  group_by(Segment, ownHome) |> 
  summarise(mean = mean(income)) |> 
  ungroup()
mu_11 <- mu_ij$mean[1]
mu_11

[1] 54497.68

segmentation |> 
  select(income, Segment, ownHome) |> 
  head(n=5)

# A tibble: 5 × 3
  income Segment    ownHome
   <dbl> <chr>      <chr>  
1 49483. Suburb mix ownNo  
2 35546. Suburb mix ownYes 
3 44169. Suburb mix ownYes 
4 81042. Suburb mix ownNo  
5 79353. Suburb mix ownYes 

• Testing Multiple Group Means: Analysis of Variance (ANOVA)

\[\begin{split} income_{ijk} = & \mu + \alpha_i + \beta_j + (\alpha\beta)_{ij} + \epsilon_{ijk} \\ & \text{ where } + \epsilon_i \sim \mathcal{N}(0, \sigma^2) \\ & \text{ and } i = 1, 2, 3, 4 \\ & j = 1, 2 \\ & k = 1, \ldots n_{ij} \\ & \mu = \mu_{11} \\ & \alpha_1 = \beta_1 = 0 \\ & (\alpha\beta)_{11} = (\alpha\beta)_{12} = 0 \\ & (\alpha\beta)_{21} = (\alpha\beta)_{31} = (\alpha\beta)_{41} = 0 \\ \end{split}\]

• Testing Multiple Group Means: Analysis of Variance (ANOVA)

\[\begin{split} \widehat{income}_{ijk} = & \widehat{\mu} + \widehat{\alpha}_i + \widehat{\beta}_j + (\widehat{\alpha\beta})_{ij} + \widehat{\epsilon}_{ijk} \\ & \text{ and } i = 1, 2, 3, 4 \\ & j = 1, 2 \\ & k = 1, \ldots n_{ij} \\ & \widehat{\mu} = \widehat{\mu}_{11} \\ & \widehat{\alpha}_2, \widehat{\alpha}_3, \widehat{\alpha}_4 \\ & \widehat{\beta}_2 \\ & (\widehat{\alpha\beta})_{22}, (\widehat{\alpha\beta})_{32}, (\widehat{\alpha\beta})_{42} \end{split}\]

\[income_{ijk} - \widehat{income}_{ijk} = \widehat{\epsilon}_{ijk}\]

• Testing Multiple Group Means: Analysis of Variance (ANOVA)

segmentation |> 
  select(income, Segment, ownHome) |> 
  head(n=2) |> 
  glimpse()

Rows: 2
Columns: 3
$ income  <dbl> 49482.81, 35546.29
$ Segment <chr> "Suburb mix", "Suburb mix"
$ ownHome <chr> "ownNo", "ownYes"

framed <- model_frame(formula = income ~ 
                                Segment + 
                                ownHome + 
                                Segment:ownHome,
            data = segmentation)

model_matrix(terms = framed$terms,
             data = framed$data) |> 
  head(n = 2) |> 
  glimpse()

Rows: 2
Columns: 8
$ `(Intercept)`                     <dbl> 1, 1
$ `SegmentSuburb mix`               <dbl> 1, 1
$ SegmentTravelers                  <dbl> 0, 0
$ `SegmentUrban hip`                <dbl> 0, 0
$ ownHomeownYes                     <dbl> 0, 1
$ `SegmentSuburb mix:ownHomeownYes` <dbl> 0, 1
$ `SegmentTravelers:ownHomeownYes`  <dbl> 0, 0
$ `SegmentUrban hip:ownHomeownYes`  <dbl> 0, 0

• Testing Multiple Group Means: Analysis of Variance (ANOVA)

  • Model

    \[\begin{bmatrix} 49482.81 \\ 35546.29 \\ \vdots \end{bmatrix} = \begin{bmatrix} 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 & 1 & 1 & 0 & 0 \\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \end{bmatrix} \begin{bmatrix} \mu \\ \alpha_2 \\ \beta_2 \\ (\alpha\beta)_{13} \\ (\alpha\beta)_{14} \\ (\alpha\beta)_{22} \\ (\alpha\beta)_{32} \\ (\alpha\beta)_{42} \\ \end{bmatrix}\]

  • Coefficients to estimate using aov

    \[\widehat{\mu} = \widehat{\mu}_{11}, \widehat{\alpha}_2, \widehat{\alpha}_3, \widehat{\alpha}_4, \widehat{\beta}_2, (\widehat{\alpha\beta})_{22}, (\widehat{\alpha\beta})_{32}, (\widehat{\alpha\beta})_{42}\]

• Testing Multiple Group Means: Analysis of Variance (ANOVA)

  • Segment

    \(H_0: \mu_{Moving\;up} = \mu_{Suburb\;mix} = \mu_{Travelers} = \mu_{Urban\;hip}\)

    \(H_1: \text{At least one group mean is different from the rest}\)

  • ownHome

    \(H_0: \mu_{ownNo} = \mu_{ownYes}\)

    \(H_1: \text{At least one group mean is different from the rest}\)

  • Segment, ownHome

    \(\begin{align} H_0: & \mu_{Moving\;up,\;ownNo} - \mu_{Moving\;up,\;ownYes} = \\ & \mu_{Suburb\;mix,\;ownNo} - \mu_{Suburb\;mix,\;ownYes} = \\ & \mu_{Travelers,\;ownNo} - \mu_{Travelers,\;ownYes} = \\ & \mu_{Urban\;hip,\;ownNo} - \mu_{Urban\;hip,\;ownYes} \end{align}\)

    \(H_1: \text{At least one difference group mean is different from the rest}\)

• Testing Multiple Group Means: Analysis of Variance (ANOVA)

model_aov <- lm(formula = income ~ -1 + Segment,
                 data = segmentation) |> 
  tidy(conf.int = TRUE)